# (Sin theta - cos theta +1)/(sin theta+ cos theta -1) =1 / (sec theta - tan theta)?

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dk_ch Share
Oct 12, 2017

$L H S = \frac{S \in \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$

$= \frac{\cos \theta \left(\sin \theta - \cos \theta + 1\right)}{\cos \theta \left(\sin \theta + \cos \theta - 1\right)}$

$= \frac{\cos \theta \left(\sin \theta - \cos \theta + 1\right)}{\cos \theta \sin \theta + {\cos}^{2} \theta - \cos \theta}$

$= \frac{\cos \theta \left(\sin \theta - \cos \theta + 1\right)}{\cos \theta \sin \theta - \cos \theta + 1 - {\sin}^{2} \theta}$

$= \frac{\cos \theta \left(\sin \theta - \cos \theta + 1\right)}{1 - {\sin}^{2} \theta - \cos \theta + \sin \theta \cos \theta}$

=(costheta(sintheta - cos theta +1))/((1- sin theta )(1+sintheta)-costheta(1-sintheta)

=(costheta(sintheta - cos theta +1))/((1- sin theta )(1+sintheta-costheta)

$= \cos \frac{\theta}{1 - \sin \theta}$

$= \frac{\cos \frac{\theta}{\cos} \theta}{\frac{1 - \sin \theta}{\cos} \theta}$

$= \frac{1}{\frac{1}{\cos} \theta - \sin \frac{\theta}{\cos} \theta}$

$= \frac{1}{\sec \theta - \tan \theta} = R H S$

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